Differential Equations – JEE Main MathematicsPractice Questions & PYQs

The differential equation $(1-x^2) dy = (xy + (x^{3+}2) \sqrt{1-x^2}) dx$ can be rearranged into the standard linear form $\frac{dy}{dx} - \frac{x}{1-x^2} y = \frac{x^{3+}2}{\sqrt{1-x^2}}$. Identifying the integrating factor involves calculating $e^{\int -\frac{x}{1-x^2} dx}$, which yields $\sqrt{1-x^2}$, and multiplying this across the entire equation allows the left side to be expressed as the derivative of the product $y \sqrt{1-x^2}$. Integrating both sides with respect to $x$ leads to $y \sqrt{1-x^2} = \int (x^{3+}2) dx$, resulting in $y \sqrt{1-x^2} = \frac{x^4}{4} + 2x + c$. Applying the condition $y(0)=0$ reveals that the constant $c$ is zero. Evaluating the target integral $\int_{-1/2}^{1/2} \sqrt{1-x^2} y dx$ therefore simplifies to calculating $\int_{-1/2}^{1/2} (\frac{x^4}{4} + 2x) dx$. Since $2x$ is an odd function, its integral over the symmetric interval $[-1/2, 1/2]$ is zero, leaving only the contribution from $\int_{-1/2}^{1/2} \frac{x^4}{4} dx$. Calculating this as $2 \int_{0}^{1/2} \frac{x^4}{4} dx$ gives \frac{1}{2} \[\frac{x^5}{5}]_0^{1/2} = \frac{1}{320}$, so$k = \frac{1}{320}$, and the final value of$k^{-1}$ is 320.

Recognizing the structure of this differential equation involves identifying that it can be rewritten to isolate a perfect differential, specifically one related to the quotient $\frac{x}{y^2}$. Starting with $2y e^{\frac{x}{y^2}} dx + (y^2 - 4x e^{\frac{x}{y^2}}) dy = 0$, we regroup the terms to isolate the exponential parts, which leads to $2e^{\frac{x}{y^2}}(y dx - 2x dy) = -y^2 dy$. Dividing both sides by $y^3$ highlights a form suitable for direct integration, as the expression $\frac{y dx - 2x dy}{y^3}$ is exactly the differential $d\left(\frac{x}{y^2}\right)$.

The equation now reads $2e^{\frac{x}{y^2}} d\left(\frac{x}{y^2}\right) = -\frac{1}{y} dy$. Integrating both sides yields $2e^{\frac{x}{y^2}} = -\ln y + c$. Applying the initial condition $x(1) = 0$ allows us to determine the constant of integration, where $2e^0 = -\ln(1) + c$, simplifying to $2 = 0 + c$, so $c = 2$.

With the constant identified, the equation becomes $2e^{\frac{x}{y^2}} = 2 - \ln y$. Calculating the value of $x$ when $y = e$ requires substituting $e$ into this equation, leading to $2e^{\frac{x}{e^2}} = 2 - \ln e$, which simplifies to $2e^{\frac{x}{e^2}} = 1$. Rearranging this result gives $e^{\frac{x}{e^2}} = \frac{1}{2}$. Taking the natural logarithm of both sides results in $\frac{x}{e^2} = \ln\left(\frac{1}{2}\right) = -\ln 2$, which means $x = -e^2 \ln 2$.

The fundamental relationship between integrals and derivatives allows the unknown function $g(x)$ to be isolated by differentiating both sides of the given equation with respect to $x$. Differentiating the integral side simply returns the integrand, while applying the quotient rule to the right side produces the expression $\frac{(e^x+1)(g(x)+x g'(x)) - x g(x) e^x}{(e^x+1)^2}$. Upon equating these two sides and multiplying by $(e^x+1)^2$, the terms containing $g(x)$ simplify directly, leaving $(e^x+1)x(\cos x-\sin x) = x(e^x+1)g'(x)$. Dividing both sides by $x(e^x+1)$ yields $g'(x) = \cos x - \sin x$, and integrating this result gives $g(x) = \sin x + \cos x + C$.

Monotonicity is determined by examining the sign of the derivative of the specific expression in question. For the combination $g(x)-g'(x)$, substituting the derived forms gives $(\sin x + \cos x + C) - (\cos x - \sin x)$, which simplifies to $2 \sin x + C$. The derivative of this expression is $2 \cos x$, which is strictly positive for all $x$ in the interval $(0, \pi/2)$, confirming that $g-g'$ is an increasing function on that interval. Conversely, the sum $g(x)+g'(x)$ simplifies to $2 \cos x + C$, which has a derivative of $-2 \sin x$; because this derivative is negative for all $x$ in $(0, \pi/2)$, the sum is decreasing.

Separating variables allows us to transform the given differential equation into the form $\frac{dy}{1+y^2} = -\frac{2e^x}{1+e^{2x}}dx$, which is readily integrable. Integrating both sides yields $\tan^{-1}(y) + 2\tan^{-1}(e^x) = C$, and applying the condition $y(0)=0$ determines that $C = \frac{\pi}{2}$. Therefore, the implicit solution is $\tan^{-1}(y) + 2\tan^{-1}(e^x) = \frac{\pi}{2}$.

Finding $y'(0)$ is straightforward by substituting $x=0$ and $y=0$ into the original differential equation $(1+e^{2x})\frac{dy}{dx}+2(1+y^2)e^x=0$, which results in $2y'(0) + 2 = 0$, giving $y'(0) = -1$. For the term $y(\ln \sqrt{3})$, substituting $x = \ln \sqrt{3}$ into our implicit solution results in $\tan^{-1}(y) + 2\tan^{-1}(\sqrt{3}) = \frac{\pi}{2}$, leading to $\tan^{-1}(y) = -\frac{\pi}{6}$ and $y = -\frac{1}{\sqrt{3}}$. Combining these results in the final expression $6(y'(0) + (y(\ln \sqrt{3}))^2) = 6(-1 + (-\frac{1}{\sqrt{3}})^2) = 6(-1 + \frac{1}{3}) = -4$.

The differential equation follows the linear form $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = \frac{\sqrt{2}}{2\cos^4 x - \cos 2x}$ and $Q(x) = x e^{\tan^{-1}(\sqrt{2}\cot 2x)}$. Defining the integrating factor as $e^{\int P(x) dx}$, recognize that the identity $2\cos^4 x - \cos 2x = \cos^4 x + \sin^4 x$ facilitates the integration. Expressing $\int \frac{\sqrt{2} \, dx}{\cos^4 x + \sin^4 x}$ by dividing the numerator and denominator by $\cos^4 x$ leads to the integral of $\frac{\sqrt{2} \sec^4 x}{1 + \tan^4 x}$, which simplifies to $-\tan^{-1}(\sqrt{2}\cot 2x)$. This makes the integrating factor $e^{-\tan^{-1}(\sqrt{2}\cot 2x)}$.

The standard solution $y \cdot \text{IF} = \int Q(x) \cdot \text{IF} \, dx + C$ yields:
$y e^{-\tan^{-1}(\sqrt{2}\cot 2x)} = \int x e^{\tan^{-1}(\sqrt{2}\cot 2x)} \cdot e^{-\tan^{-1}(\sqrt{2}\cot 2x)} \, dx = \int x \, dx = \frac{x^2}{2} + C$
Substituting the boundary condition $y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{32}$, note that $\cot(2 \cdot \frac{\pi}{4}) = \cot(\frac{\pi}{2}) = 0$ and $e^{-\tan^{-1}(0)} = 1$. The equation $\frac{\pi^2}{32} = \frac{(\pi/4)^2}{2} + C$ reveals that $C = 0$.

With the specific solution $y = \frac{x^2}{2} e^{\tan^{-1}(\sqrt{2}\cot 2x)}$, evaluating at $x = \frac{\pi}{3}$ gives:
$y\left(\frac{\pi}{3}\right) = \frac{(\pi/3)^2}{2} e^{\tan^{-1}(\sqrt{2}\cot(2\pi/3))} = \frac{\pi^2}{18} e^{\tan^{-1}(-\sqrt{2/3})}$
Applying the property $\tan^{-1}(-z) = -\tan^{-1}(z)$, this expression matches $\frac{\pi^2}{18} e^{-\tan^{-1}(\sqrt{2/3})}$. Equating this to the provided form $\frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}$ identifies $\alpha = \sqrt{2/3}$, thus $3\alpha^2 = 3\left(\frac{2}{3}\right) = 2$.

The differential equation $\frac{dy}{dx}+e^x(x^{2-}2)y=(x^{2-}2x)(x^{2-}2)e^{2x}$ is a first-order linear equation, which can be solved using the integrating factor $e^{\int e^x(x^{2-}2)dx} = e^{(x^{2-}2x)e^x}$. Multiplying the entire equation by this factor allows the left side to be expressed as $\frac{d}{dx}(y \cdot e^{(x^{2-}2x)e^x})$, while the right side becomes the integral of $(x^{2-}2x)e^x \cdot (x^{2-}2)e^x \cdot e^{(x^{2-}2x)e^x} \, dx$. Using the substitution $t=(x^{2-}2x)e^x$ and $dt=(x^{2-}2)e^x \, dx$, the integral simplifies to $\int t e^t \, dt = (t-1)e^t+c$, which leads to the general solution $y = (x^{2-}2x)e^x - 1 + c e^{-(x^{2-}2x)e^x}$. Applying the condition $y(0)=0$ fixes $c=1$, and evaluating the expression at $x=2$ gives $y = (2^{2-}2(2))e^2 - 1 + 1 \cdot e^{-(2^{2-}2(2))e^2} = 0 - 1 + 1 = 0$.

Rearranging the given differential equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ allows us to identify the coefficient $P(x)$ as $\frac{4x}{\sin(2x^2) \ln(\tan x^2)}$. By substituting $u = \tan x^2$, the integral of $P(x)$ simplifies into a straightforward logarithmic form, identifying the integrating factor as $\ln(\tan x^2)$. Multiplying the differential equation by this factor transforms the left side into the derivative of the product $y \ln(\tan x^2)$.

The right side of the equation is then integrated by applying trigonometric identities to the numerator $\sin(x^2 - \pi/4)$, which leads to an expression involving the integrals of secant and cosecant functions. This integration yields the general solution $y \ln(\tan x^2) = \ln \left| \frac{\sec x^2 + \tan x^2}{\csc x^2 - \cot x^2} \right| + C$.

Applying the condition that the curve passes through the point $(\sqrt{\pi/6}, 1)$ allows for the calculation of the constant $C$. Once this constant is determined, we evaluate the expression for $y$ at $x = \sqrt{\pi/3}$. Given the logarithmic values at this point, the equation simplifies to $y \ln\sqrt{3} = \ln(1/\sqrt{3})$, which solves to $y = -1$. Therefore, the absolute value $|y(\sqrt{\pi/3})|$ is 1.

Solving linear differential equations often relies on identifying an appropriate integrating factor to transform the equation into a derivative of a product. In this case, rearranging the equation $\frac{dy}{dx} + \frac{2}{x}y = e^x$ allows us to find the integrating factor $e^{\int \frac{2}{x} dx} = x^2$. Multiplying the differential equation by $x^2$ yields $\frac{d}{dx}(x^2 y) = x^2 e^x$, and integrating both sides results in $x^2 y = e^x(x^2 - 2x + 2) + c$. Applying the condition $y(1)=0$ reveals that $0 = e + c$, so the constant of integration is $c = -e$.

Given the function $z(x) = x^2 y - e^x$, we substitute the expression for $x^2 y$ to obtain $z(x) = e^x(x^2 - 2x + 2) - e - e^x$, which simplifies to $z(x) = e^x(x-1)^2 - e$. To identify the local maximum, we find the critical points by setting the derivative $z'(x) = e^x(x-1)^2 + 2e^x(x-1)$ to zero. Factoring the expression gives $z'(x) = e^x(x-1)(x+1)$, which indicates stationary points at $x=1$ and $x=-1$. Evaluating the original function $z(x)$ at $x=-1$ confirms the local maximum value is $\frac{4}{e} - e$.

This problem involves a homogeneous differential equation because every term in the numerator and denominator has a total degree of three, which justifies the substitution $y = vx$ where $v$ is a function of $x$. Differentiating $y = vx$ yields $\frac{dy}{dx} = v + x \frac{dv}{dx}$, and substituting this into the original equation results in $v + x \frac{dv}{dx} = \frac{4v^{3+}2v}{3v^{2+}1}$. Rearranging to isolate the variables, we obtain the integral $\int \frac{3v^{2+}1}{v^{3+}v} dv = \int \frac{dx}{x}$, which evaluates to $\ln |v^{3+}v| = \ln x + C$. Substituting $v = \frac{y}{x}$ back into the result gives the implicit relation $\ln |(\frac{y}{x})^3 + \frac{y}{x}| = \ln x + C$, and applying the initial condition $y(1)=1$ determines that $C = \ln 2$. Evaluating this at $x=2$ leads to the equation $\frac{y^3}{8} + \frac{y}{2} = 4$, which simplifies to $y^3 + 4y = 32$. Since the expression $y^3 + 4y$ equals $16$ when $y=2$ and $39$ when $y=3$, the value $y(2)$ must fall between 2 and 3. Given $y(2) \in [n-1, n)$, this interval is $[2, 3)$, identifying $n$ as 3.

Separating the variables in the equation $\frac{dy}{dx} + \frac{2^{x-y}(2^y-1)}{2^x-1} = 0$ begins by expressing $2^{x-y}$ as $\frac{2^x}{2^y}$, leading to $\frac{dy}{dx} = -\frac{2^x(2^y-1)}{2^y(2^x-1)}$. Grouping terms with $y$ and $x$ separately results in the integration of $\frac{2^y}{2^y-1} \, dy = -\frac{2^x}{2^x-1} \, dx$. Applying integration yields $\log_2(2^y-1) = -\log_2(2^x-1) + C$, which consolidates into the product relationship $(2^y-1)(2^x-1) = K$.

Using the initial condition $y(1)=1$ allows us to find the constant, as $(2^{1-}1)(2^{1-}1) = K$ results in $K=1$. Substituting $x=2$ into the established relationship $(2^y-1)(2^x-1) = 1$ gives $(2^y-1)(2^{2-}1) = 1$, or $(2^y-1)(3) = 1$. Solving for $2^y$ yields $2^y-1 = \frac{1}{3}$, so $2^y = \frac{4}{3}$. Applying a base-2 logarithm to both sides results in $y = \log_2\left(\frac{4}{3}\right)$, which simplifies to $2 - \log_2 3$.

A system of circles passing through two fixed points can be represented by a base circle equation modified by a parameter multiplied by the line passing through those two points. Since the points $(0, 2)$ and $(0, -2)$ lie on the y-axis, the line connecting them is $x = 0$. Consequently, the entire family of circles passing through these points is defined by the equation $x^2 + y^2 + \lambda x - 4 = 0$, where $\lambda$ is an arbitrary constant that represents the varying horizontal position of the center for each circle in the family.

To derive the differential equation, we differentiate this family with respect to $x$ to eliminate the parameter $\lambda$. This gives $2x + 2y \frac{dy}{dx} + \lambda = 0$, which allows us to isolate the parameter as $\lambda = -(2x + 2y \frac{dy}{dx})$. Substituting this expression back into the original family equation yields $x^2 + y^2 - x(2x + 2y \frac{dy}{dx}) - 4 = 0$. Expanding the terms results in $x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} - 4 = 0$, which simplifies to $-x^2 + y^2 - 2xy \frac{dy}{dx} - 4 = 0$. Multiplying this entire result by $-1$ provides the final form $2xy \frac{dy}{dx} + x^2 - y^2 + 4 = 0$.

Recognizing the homogeneous structure of the differential equation, the substitution $v = \frac{y}{x}$ simplifies the expression significantly. Substituting $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the original equation reduces it to $v^2x + x^2v \frac{dv}{dx} = v^2x + \frac{x \phi(v^2)}{\phi'(v^2)}$, which simplifies further to the separable form $x v \frac{dv}{dx} = \frac{\phi(v^2)}{\phi'(v^2)}$. Rearranging the variables leads to the relation $\frac{v \phi'(v^2)}{\phi(v^2)} dv = \frac{dx}{x}$, allowing for independent integration on both sides.

Integrating the left side using the substitution $u = v^2$, where $du = 2v dv$, yields $\frac{1}{2} \ln(\phi(v^2)) = \ln(x) + C$. This reverts to $\frac{1}{2} \ln(\phi(\frac{y^2}{x^2})) = \ln(x) + C$ after substituting $v = \frac{y}{x}$ back into the expression. Applying the initial condition $y(1) = -1$, where $x=1$, we find the integration constant is $C = \frac{1}{2} \ln(\phi(1))$.

Substituting the constant back into the integrated form gives $\frac{1}{2} \ln(\phi(\frac{y^2}{x^2})) = \ln(x) + \frac{1}{2} \ln(\phi(1))$, which simplifies to $\ln(\phi(\frac{y^2}{x^2})) = 2\ln(x) + \ln(\phi(1))$. This expression is equivalent to $\ln(\phi(\frac{y^2}{x^2})) = \ln(x^2 \phi(1))$, or simply $\phi(\frac{y^2}{x^2}) = x^2 \phi(1)$. Evaluating this result at $x = 2$ provides the final value $\phi(\frac{y^2}{4}) = 4 \phi(1)$.

Substituting $Y = y+1$ and $X = x+2$ transforms the differential equation into $X dY - Y dX = X e^{Y/X} dX$. Dividing by $X^2$ yields the derivative of the quotient $d(Y/X) = e^{Y/X} \frac{dX}{X}$, which, upon separating variables and integrating, leads to the relation:

$-e^{-Y/X} = \ln|X| + C$

Applying the initial condition $y(1)=1$, which corresponds to $Y=2$ and $X=3$, determines the constant of integration such that the solution takes the form $e^{-Y/X} = e^{2/3} + \ln 3 - \ln|X|$. For the solution to be valid, the exponential expression must be positive, which establishes the domain constraint $\ln|X| < e^{2/3} + \ln 3$. Defining $\lambda = e^{2/3} + \ln 3$, the inequality $|x+2| < e^{\lambda}$ defines the interval limits $\alpha = -e^{\lambda} - 2$ and $\beta = e^{\lambda} - 2$. Summing these values gives $\alpha + \beta = -4$, so the final magnitude $|\alpha + \beta|$ is 4.

Both differential equations describe homogeneous families of circles. For the first equation, substituting $y=vx$ allows for the separation of variables, leading to the integral form that defines the circle equation $x^{2+}y^2=Cx$. Given that the curve passes through $(1,1)$, we calculate $1^{2+}1^2=C(1)$, which yields $C=2$ and results in the circle $x^{2+}y^2=2x$, representing a circle centered at $(1,0)$ with radius 1.

The second differential equation similarly yields the circular family $x^{2+}y^2=Cy$. Applying the condition that it passes through $(1,1)$ confirms $C=2$, defining the circle $x^{2+}y^2=2y$ centered at $(0,1)$ with radius 1. These two circles intersect at the origin $(0,0)$ and the point $(1,1)$, creating a lens-shaped region of overlap between them.

The area of this enclosed region is twice the area of the circular segment formed by the chord connecting the origin and $(1,1)$. For a unit circle, the sector area corresponding to a 90-degree angle is $\frac{\pi}{4}$, and subtracting the area of the right triangle with vertices at $(0,0)$, $(1,0)$, and $(1,1)$ yields a segment area of $\frac{\pi}{4}-\frac{1}{2}$. Doubling this segment area to account for both symmetric halves gives a total enclosed area of $\frac{\pi}{2}-1$.

The differential equation $2 x^2 dy + (e^y - 2x) dx = 0$ can be rearranged as $\frac{dy}{dx} + \frac{e^y}{2x^2} = \frac{1}{x}$. Multiplying the equation by $e^{-y}$ leads to $e^{-y} \frac{dy}{dx} - \frac{e^{-y}}{x} = -\frac{1}{2x^2}$. By substituting $t = e^{-y}$, the derivative term becomes $\frac{dt}{dx} = -e^{-y} \frac{dy}{dx}$, which transforms the equation into $-\frac{dt}{dx} - \frac{t}{x} = -\frac{1}{2x^2}$, or equivalently $\frac{dt}{dx} + \frac{t}{x} = \frac{1}{2x^2}$.

This is a first-order linear differential equation with an integrating factor equal to $e^{\int \frac{1}{x} dx} = x$. Applying this factor results in $x \frac{dt}{dx} + t = \frac{1}{2x}$, which is the derivative $\frac{d}{dx} (xt) = \frac{1}{2x}$. Integrating both sides gives $xt = \frac{1}{2} \ln x + C$, or $2x e^{-y} = \ln x + c$ where $c = 2C$. Using the condition $y(e) = 1$, we find $2e \cdot e^{-1} = \ln e + c$, which simplifies to $2 = 1 + c$, confirming $c = 1$. For $x = 1$, the equation becomes $2(1) e^{-y} = \ln 1 + 1$, yielding $2e^{-y} = 1$, so $e^y = 2$ and $y = \log_e 2$.

The behavior of the function $y(x)$ is determined by a first-order linear differential equation derived from the determinant of the matrix $A$. Expanding the determinant of $A$ along the first column yields $|A| = -\frac{y}{x} + 2 \sin x + 2$, which, when set equal to $\frac{dy}{dx}$, results in the linear differential equation $\frac{dy}{dx} + \frac{y}{x} = 2 \sin x + 2$. This equation is solved by identifying the integrating factor $e^{\int \frac{1}{x} dx} = x$ and multiplying the entire equation by this factor, leading to the derivative of the product $xy$, expressed as $\frac{d}{dx}(xy) = 2x \sin x + 2x$. Integrating both sides provides the general solution: